answer:The basis for the null space of the given matrix is left{ begin{pmatrix} 35.000 -2.000 0.000 0.000 79.000 end{pmatrix}, begin{pmatrix} 91.000 -21.000 79.000 0.000 0.000 end{pmatrix}, begin{pmatrix} 114.000 -2.000 0.000 79.000 0.000 end{pmatrix} right}.

answer:The proof relies on the definition of the characteristic of a ring and the fact that a finite ring cannot have characteristic zero. The key idea is to consider the set of all multiples of 1 in R and show that this set is infinite. If R were finite, then this set would also be finite, and there would exist two distinct natural numbers m and n such that mcdot 1 = ncdot 1. This would imply that (m-n)cdot 1 = 0, which contradicts the fact that R has characteristic zero. Therefore, R must be infinite. Let R be a unital ring of characteristic zero. We will prove that R is infinite by contradiction. Suppose that R is finite. Then there exists a positive integer n such that ncdot 1 = 0 for all n in mathbb{N}. But this contradicts the fact that R has characteristic zero. Therefore, R must be infinite.

answer:To differentiate the given function, we can use the sum rule and the chain rule. The sum rule states that the derivative of a sum of functions is equal to the sum of the derivatives of each function. The chain rule states that the derivative of a function composed with another function is equal to the product of the derivative of the outer function and the derivative of the inner function. Using these rules, we can differentiate the given function as follows: frac{d}{dx} left[sqrt[6]{-7 x-frac{5}{2}}+sin left(4 x+frac{11}{2}right)right] = frac{d}{dx} left[sqrt[6]{-7 x-frac{5}{2}}right] + frac{d}{dx} left[sin left(4 x+frac{11}{2}right)right] = frac{1}{6} left(-7 x-frac{5}{2}right)^{-5/6} cdot (-7) + 4 cos left(4 x+frac{11}{2}right) = frac{-7}{6 sqrt[6]{(-7 x-frac{5}{2})^{5/6}}}+4 cos left(4 x+frac{11}{2}right) Therefore, the derivative of the given function is frac{-7}{6 sqrt[6]{(-7 x-frac{5}{2})^{5/6}}}+4 cos left(4 x+frac{11}{2}right). The answer is frac{-7}{6 sqrt[6]{(-7 x-frac{5}{2})^{5/6}}}+4 cos left(4 x+frac{11}{2}right)

answer:The Jacobian matrix is a matrix of partial derivatives that describes the linear transformation of a vector-valued function. In this case, the Jacobian matrix of mathbf{r}(x, y, z) is a 3x3 matrix whose entries are the partial derivatives of the component functions of mathbf{r}(x, y, z). The partial derivatives of f(x, y, z), g(x, y, z), and h(x, y, z) are: frac{partial f}{partial x} = 1, quad frac{partial f}{partial y} = 0, quad frac{partial f}{partial z} = 0 frac{partial g}{partial x} = 0, quad frac{partial g}{partial y} = -frac{1}{sqrt{1-y^2}}, quad frac{partial g}{partial z} = 0 frac{partial h}{partial x} = 0, quad frac{partial h}{partial y} = frac{z}{3sqrt[3]{(yz)^2}}, quad frac{partial h}{partial z} = frac{y}{3sqrt[3]{(yz)^2}} Substituting these partial derivatives into the Jacobian matrix, we obtain the desired result. The Jacobian matrix of mathbf{r}(x, y, z) is given by: J(mathbf{r}(x, y, z)) = begin{bmatrix} frac{partial f}{partial x} & frac{partial f}{partial y} & frac{partial f}{partial z} frac{partial g}{partial x} & frac{partial g}{partial y} & frac{partial g}{partial z} frac{partial h}{partial x} & frac{partial h}{partial y} & frac{partial h}{partial z} end{bmatrix} = begin{bmatrix} 1 & 0 & 0 0 & -frac{1}{sqrt{1-y^2}} & 0 0 & frac{z}{3sqrt[3]{(yz)^2}} & frac{y}{3sqrt[3]{(yz)^2}} end{bmatrix}